# X-ray studies of silver crystals show that it forms CCP lattice and edge length of the unit cell is 408.6 pm. Calculate the density of the silver (atomic mass = 107.9 u)

X-ray studies of silver crystals show that it forms CCP lattice and edge length of the unit cell is 408.6 pm. Calculate the density of the silver (atomic mass = 107.9 u)

Given Values

Unit Cell = CCP or FCC (Z=4)

Atomic mass of Silver (M) = 107.9 ≈108 u

Edge length (a) = 408.6 pm

= 408.6 × 10-10 cm

Avogadro’s number (NA) = 6.02 × 1023

Calculations

$$Density=\frac{Z \times M}{a^{3}\times N_{A}}~~gm/cm^{3}$$

$$D=\frac{4 \times 108}{\left ( 408.6\times 10^{-10} \right )^{3}\times 6.02 \times 10^{23}}$$

$$D=\frac{432\times 10^{7}}{\left ( 408.6 \right )^{3}\times 6.02}$$ = 10.5 gm/cm3